HW12S - Gauthier, Joseph Homework 12 Due: Dec 4 2007,...

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Unformatted text preview: Gauthier, Joseph Homework 12 Due: Dec 4 2007, midnight Inst: Vandenbout 1 This print-out should have 32 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. The due time is Central time. EXAM 4 is Thursday, 12/6 from 7-9 PM Go to the right room A-K WEL 1.316 L-O WEL 2.122 P-Z WEL 1.308 001 (part 1 of 1) 10 points For a given transfer of energy, a greater change in disorder occurs when the temperature is high. 1. True 2. False correct Explanation: 002 (part 1 of 1) 10 points Entropy is a state function. 1. False 2. True correct Explanation: 003 (part 1 of 1) 10 points Place the following in order of increasing en- tropy. 1. solid, liquid, and gas correct 2. liqiud, solid, and gas 3. gas, solid, and liqiud 4. solid, gas, and liqiud 5. gas, liqiud, and solid Explanation: Entropy ( S ) is high for systems with high degrees of freedom, disorder or randomness and low for systems with low degrees of free- dom, disorder or randomness. S (g) > S ( ) > S (s) . 004 (part 1 of 1) 10 points Which substance has the higher molar en- tropy? 1. They are the same 2. Kr(g) at 298 K and 1.00 atm correct 3. Unable to determine 4. Ne(g) at 298 K and 1.00 atm Explanation: Kr(g) is more massive and has more elemen- tary particles, hence a higher molar entropy. 005 (part 1 of 1) 10 points Calculate the standard entropy of vaporiza- tion of ethanol at its boiling point 352 K. The standard molar enthalpy of vaporization of ethanol at its boiling point is 40.5 kJ mol- 1 . 1.- 40.5 kJ K- 1 mol- 1 2. +115 J K- 1 mol- 1 correct 3. +40.5 kJ K- 1 mol- 1 4.- 115 J K- 1 mol- 1 5. +513 J K- 1 mol- 1 Explanation: 006 (part 1 of 1) 10 points Assuming that the heat capacity of an ideal gas is independent of temperature, what is the entropy change associated with lowering the temperature of 3 . 1 mol of ideal gas atoms from 108 . 33 C to- 41 . 65 C at constant volume? Correct answer:- 19 . 3197 J / K. Explanation: T 1 = 108 . 33 C + 273 = 381 . 33 K T 2 =- 41 . 65 C + 273 = 231 . 35 K Gauthier, Joseph Homework 12 Due: Dec 4 2007, midnight Inst: Vandenbout 2 n = 3 . 1 mol R = 8 . 314 J K mol At constant volume, dq = n C V dT , so dS = dq T = n C V dT T Z dS = n C V Z dT T S = n C V ln T 2 T 1 For an ideal monatomic gas C V = 3 2 R , so S = (3 . 1 mol) 3 2 8 . 314 J K mol ln 231 . 35 K 381 . 33 K =- 19 . 3197 J / K . 007 (part 1 of 1) 10 points What is the entropy change associated with the isothermal compression of 7 . 39 mol of ideal gas atoms from 6 . 27 atm to 15 . 1 atm? Correct answer:- 54 . 0012 J / K. Explanation: P 1 = 6 . 27 atm P 2 = 15 . 1 atm n = 7 . 39 mol R = 8 . 314 J K mol Because the process is isothermal, U = 0, so q =- w , where w =- P dV ....
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HW12S - Gauthier, Joseph Homework 12 Due: Dec 4 2007,...

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